Using SUM() without grouping the results

Question

I already read (this), but couldn\'t figure out a way to implement it to my specific problem. I know SUM() is an aggregate function and it doesn\'t make sense n

Answer 1

This does just one sum() query, so it should perform OK:

SELECT a.id, b.amount
FROM table1 a
cross join (SELECT SUM(amount) as amount FROM table1 AS amount) b

Answer 2

With MS SQL you can use OVER()

 select id, SUM(amount) OVER()
 from table1;

select id, SUM(amount) OVER()
from (
  select 1 as id, 23 as amount
  union all
  select 2 as id, 11 as amount
  union all
  select 3 as id, 8 as amount
  union all
  select 4 as id, 7 as amount
) A

--- OVER PARTITION ID

PARTITION BY which is very useful when you want to do SUM() per MONTH for example or do quarterly reports sales or yearly... (Note needs distinct it is doing for all rows)

 select distinct id, SUM(amount) OVER(PARTITION BY id) as [SUM_forPARTITION]
 from (
     select 1 as id, 23 as amount
     union all
     select 1 as id, 23 as amount
     union all
     select 2 as id, 11 as amount
     union all
     select 2 as id, 11 as amount
     union all
     select 3 as id, 8 as amount
     union all
     select 4 as id, 7 as amount
) OverPARTITIONID

Answer 3

Join the original table to the sum with a subquery:

SELECT * FROM table1, (SELECT SUM(amount) FROM table1 AS amount) t

Answer 4

SELECT a.id, b.amount
FROM table1 a
CROSS JOIN
(
    SELECT SUM(amount) amount FROM table1
) b

You need to perform a cartesian join of the value of the sum of every row in the table to each id. Since there is only one result of the subselect (49), it basically just gets tacked onto each id.